"""Some simple financial calculations

patterned after spreadsheet computations.

There is some complexity in each function
so that the functions behave like ufuncs with
broadcasting and being able to be called with scalars
or arrays (or other sequences).

Functions support the :class:`decimal.Decimal` type unless
otherwise stated.
"""
from __future__ import division, absolute_import, print_function

from decimal import Decimal

import numpy as np


__all__ = ['fv', 'pmt', 'nper', 'ipmt', 'ppmt', 'pv', 'rate',
           'irr', 'npv', 'mirr']

_when_to_num = {'end': 0, 'begin': 1,
                'e': 0, 'b': 1,
                0: 0, 1: 1,
                'beginning': 1,
                'start': 1,
                'finish': 0}

class npf(object):

    @staticmethod
    def _convert_when(when):
        # Test to see if when has already been converted to ndarray
        # This will happen if one function calls another, for example ppmt
        if isinstance(when, np.ndarray):
            return when
        try:
            return _when_to_num[when]
        except (KeyError, TypeError):
            return [_when_to_num[x] for x in when]


    @staticmethod
    def fv(rate, nper, pmt, pv, when='end'):
        """
        Compute the future value.

        Given:
         * a present value, `pv`
         * an interest `rate` compounded once per period, of which
           there are
         * `nper` total
         * a (fixed) payment, `pmt`, paid either
         * at the beginning (`when` = {'begin', 1}) or the end
           (`when` = {'end', 0}) of each period

        Return:
           the value at the end of the `nper` periods

        Parameters
        ----------
        rate : scalar or array_like of shape(M, )
            Rate of interest as decimal (not per cent) per period
        nper : scalar or array_like of shape(M, )
            Number of compounding periods
        pmt : scalar or array_like of shape(M, )
            Payment
        pv : scalar or array_like of shape(M, )
            Present value
        when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
            When payments are due ('begin' (1) or 'end' (0)).
            Defaults to {'end', 0}.

        Returns
        -------
        out : ndarray
            Future values.  If all input is scalar, returns a scalar float.  If
            any input is array_like, returns future values for each input element.
            If multiple inputs are array_like, they all must have the same shape.

        Notes
        -----
        The future value is computed by solving the equation::

         fv +
         pv*(1+rate)**nper +
         pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0

        or, when ``rate == 0``::

         fv + pv + pmt * nper == 0

        References
        ----------
        .. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
           Open Document Format for Office Applications (OpenDocument)v1.2,
           Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
           Pre-Draft 12. Organization for the Advancement of Structured Information
           Standards (OASIS). Billerica, MA, USA. [ODT Document].
           Available:
           http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
           OpenDocument-formula-20090508.odt

        Examples
        --------
        >>> import numpy as np
        >>> import numpy_financial as npf

        What is the future value after 10 years of saving $100 now, with
        an additional monthly savings of $100.  Assume the interest rate is
        5% (annually) compounded monthly?

        >>> npf.fv(0.05/12, 10*12, -100, -100)
        15692.928894335748

        By convention, the negative sign represents cash flow out (i.e. money not
        available today).  Thus, saving $100 a month at 5% annual interest leads
        to $15,692.93 available to spend in 10 years.

        If any input is array_like, returns an array of equal shape.  Let's
        compare different interest rates from the example above.

        >>> a = np.array((0.05, 0.06, 0.07))/12
        >>> npf.fv(a, 10*12, -100, -100)
        array([ 15692.92889434,  16569.87435405,  17509.44688102]) # may vary

        """
        when = npf._convert_when(when)
        rate, nper, pmt, pv, when = np.broadcast_arrays(rate, nper, pmt, pv, when)

        fv_array = np.empty_like(rate)
        zero = rate == 0
        nonzero = ~zero

        fv_array[zero] = -(pv[zero] + pmt[zero] * nper[zero])

        rate_nonzero = rate[nonzero]
        temp = (1 + rate_nonzero)**nper[nonzero]
        fv_array[nonzero] = (
            - pv[nonzero] * temp
            - pmt[nonzero] * (1 + rate_nonzero * when[nonzero]) / rate_nonzero
            * (temp - 1)
        )

        if np.ndim(fv_array) == 0:
            # Follow the ufunc convention of returning scalars for scalar
            # and 0d array inputs.
            return fv_array.item(0)
        return fv_array

    @staticmethod
    def pmt(rate, nper, pv, fv=0, when='end'):
        """
        Compute the payment against loan principal plus interest.

        Given:
         * a present value, `pv` (e.g., an amount borrowed)
         * a future value, `fv` (e.g., 0)
         * an interest `rate` compounded once per period, of which
           there are
         * `nper` total
         * and (optional) specification of whether payment is made
           at the beginning (`when` = {'begin', 1}) or the end
           (`when` = {'end', 0}) of each period

        Return:
           the (fixed) periodic payment.

        Parameters
        ----------
        rate : array_like
            Rate of interest (per period)
        nper : array_like
            Number of compounding periods
        pv : array_like
            Present value
        fv : array_like,  optional
            Future value (default = 0)
        when : {{'begin', 1}, {'end', 0}}, {string, int}
            When payments are due ('begin' (1) or 'end' (0))

        Returns
        -------
        out : ndarray
            Payment against loan plus interest.  If all input is scalar, returns a
            scalar float.  If any input is array_like, returns payment for each
            input element. If multiple inputs are array_like, they all must have
            the same shape.

        Notes
        -----
        The payment is computed by solving the equation::

         fv +
         pv*(1 + rate)**nper +
         pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0

        or, when ``rate == 0``::

          fv + pv + pmt * nper == 0

        for ``pmt``.

        Note that computing a monthly mortgage payment is only
        one use for this function.  For example, pmt returns the
        periodic deposit one must make to achieve a specified
        future balance given an initial deposit, a fixed,
        periodically compounded interest rate, and the total
        number of periods.

        References
        ----------
        .. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
           Open Document Format for Office Applications (OpenDocument)v1.2,
           Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
           Pre-Draft 12. Organization for the Advancement of Structured Information
           Standards (OASIS). Billerica, MA, USA. [ODT Document].
           Available:
           http://www.oasis-open.org/committees/documents.php
           ?wg_abbrev=office-formulaOpenDocument-formula-20090508.odt

        Examples
        --------
        >>> import numpy_financial as npf

        What is the monthly payment needed to pay off a $200,000 loan in 15
        years at an annual interest rate of 7.5%?

        >>> npf.pmt(0.075/12, 12*15, 200000)
        -1854.0247200054619

        In order to pay-off (i.e., have a future-value of 0) the $200,000 obtained
        today, a monthly payment of $1,854.02 would be required.  Note that this
        example illustrates usage of `fv` having a default value of 0.

        """
        when = npf._convert_when(when)
        (rate, nper, pv, fv, when) = map(np.array, [rate, nper, pv, fv, when])
        temp = (1 + rate)**nper
        mask = (rate == 0)
        masked_rate = np.where(mask, 1, rate)
        fact = np.where(mask != 0, nper,
                        (1 + masked_rate*when)*(temp - 1)/masked_rate)
        return -(fv + pv*temp) / fact

    @staticmethod
    def nper(rate, pmt, pv, fv=0, when='end'):
        """
        Compute the number of periodic payments.

        :class:`decimal.Decimal` type is not supported.

        Parameters
        ----------
        rate : array_like
            Rate of interest (per period)
        pmt : array_like
            Payment
        pv : array_like
            Present value
        fv : array_like, optional
            Future value
        when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
            When payments are due ('begin' (1) or 'end' (0))

        Notes
        -----
        The number of periods ``nper`` is computed by solving the equation::

         fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate*((1+rate)**nper-1) = 0

        but if ``rate = 0`` then::

         fv + pv + pmt*nper = 0

        Examples
        --------
        >>> import numpy as np
        >>> import numpy_financial as npf

        If you only had $150/month to pay towards the loan, how long would it take
        to pay-off a loan of $8,000 at 7% annual interest?

        >>> print(np.round(npf.nper(0.07/12, -150, 8000), 5))
        64.07335

        So, over 64 months would be required to pay off the loan.

        The same analysis could be done with several different interest rates
        and/or payments and/or total amounts to produce an entire table.

        >>> npf.nper(*(np.ogrid[0.07/12: 0.08/12: 0.01/12,
        ...                     -150   : -99    : 50    ,
        ...                     8000   : 9001   : 1000]))
        array([[[ 64.07334877,  74.06368256],
                [108.07548412, 127.99022654]],
               [[ 66.12443902,  76.87897353],
                [114.70165583, 137.90124779]]])

        """
        when = npf._convert_when(when)
        rate, pmt, pv, fv, when = np.broadcast_arrays(rate, pmt, pv, fv, when)
        nper_array = np.empty_like(rate, dtype=np.float64)

        zero = rate == 0
        nonzero = ~zero

        with np.errstate(divide='ignore'):
            # Infinite numbers of payments are okay, so ignore the
            # potential divide by zero.
            nper_array[zero] = -(fv[zero] + pv[zero]) / pmt[zero]

        nonzero_rate = rate[nonzero]
        z = pmt[nonzero] * (1 + nonzero_rate * when[nonzero]) / nonzero_rate
        nper_array[nonzero] = (
            np.log((-fv[nonzero] + z) / (pv[nonzero] + z))
            / np.log(1 + nonzero_rate)
        )

        return nper_array

    @staticmethod
    def _value_like(arr, value):
        entry = arr.item(0)
        if isinstance(entry, Decimal):
            return Decimal(value)
        else:
            return np.array(value, dtype=arr.dtype).item(0)

    @staticmethod
    def ipmt(rate, per, nper, pv, fv=0, when='end'):
        """
        Compute the interest portion of a payment.

        Parameters
        ----------
        rate : scalar or array_like of shape(M, )
            Rate of interest as decimal (not per cent) per period
        per : scalar or array_like of shape(M, )
            Interest paid against the loan changes during the life or the loan.
            The `per` is the payment period to calculate the interest amount.
        nper : scalar or array_like of shape(M, )
            Number of compounding periods
        pv : scalar or array_like of shape(M, )
            Present value
        fv : scalar or array_like of shape(M, ), optional
            Future value
        when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
            When payments are due ('begin' (1) or 'end' (0)).
            Defaults to {'end', 0}.

        Returns
        -------
        out : ndarray
            Interest portion of payment.  If all input is scalar, returns a scalar
            float.  If any input is array_like, returns interest payment for each
            input element. If multiple inputs are array_like, they all must have
            the same shape.

        See Also
        --------
        ppmt, pmt, pv

        Notes
        -----
        The total payment is made up of payment against principal plus interest.

        ``pmt = ppmt + ipmt``

        Examples
        --------
        >>> import numpy as np
        >>> import numpy_financial as npf

        What is the amortization schedule for a 1 year loan of $2500 at
        8.24% interest per year compounded monthly?

        >>> principal = 2500.00

        The 'per' variable represents the periods of the loan.  Remember that
        financial equations start the period count at 1!

        >>> per = np.arange(1*12) + 1
        >>> ipmt = npf.ipmt(0.0824/12, per, 1*12, principal)
        >>> ppmt = npf.ppmt(0.0824/12, per, 1*12, principal)

        Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal
        'pmt'.

        >>> pmt = npf.pmt(0.0824/12, 1*12, principal)
        >>> np.allclose(ipmt + ppmt, pmt)
        True

        >>> fmt = '{0:2d} {1:8.2f} {2:8.2f} {3:8.2f}'
        >>> for payment in per:
        ...     index = payment - 1
        ...     principal = principal + ppmt[index]
        ...     print(fmt.format(payment, ppmt[index], ipmt[index], principal))
         1  -200.58   -17.17  2299.42
         2  -201.96   -15.79  2097.46
         3  -203.35   -14.40  1894.11
         4  -204.74   -13.01  1689.37
         5  -206.15   -11.60  1483.22
         6  -207.56   -10.18  1275.66
         7  -208.99    -8.76  1066.67
         8  -210.42    -7.32   856.25
         9  -211.87    -5.88   644.38
        10  -213.32    -4.42   431.05
        11  -214.79    -2.96   216.26
        12  -216.26    -1.49    -0.00

        >>> interestpd = np.sum(ipmt)
        >>> np.round(interestpd, 2)
        -112.98

        """
        when = npf._convert_when(when)
        rate, per, nper, pv, fv, when = np.broadcast_arrays(rate, per, nper,
                                                            pv, fv, when)

        total_pmt = npf.pmt(rate, nper, pv, fv, when)
        ipmt_array = np.array(npf._rbl(rate, per, total_pmt, pv, when) * rate)

        # Payments start at the first period, so payments before that
        # don't make any sense.
        ipmt_array[per < 1] = npf._value_like(ipmt_array, np.nan)
        # If payments occur at the beginning of a period and this is the
        # first period, then no interest has accrued.
        per1_and_begin = (when == 1) & (per == 1)
        ipmt_array[per1_and_begin] = npf._value_like(ipmt_array, 0)
        # If paying at the beginning we need to discount by one period.
        per_gt_1_and_begin = (when == 1) & (per > 1)
        ipmt_array[per_gt_1_and_begin] = (
            ipmt_array[per_gt_1_and_begin] / (1 + rate[per_gt_1_and_begin])
        )

        if np.ndim(ipmt_array) == 0:
            # Follow the ufunc convention of returning scalars for scalar
            # and 0d array inputs.
            return ipmt_array.item(0)
        return ipmt_array

    @staticmethod
    def _rbl(rate, per, pmt, pv, when):
        """
        This function is here to simply have a different name for the 'fv'
        function to not interfere with the 'fv' keyword argument within the 'ipmt'
        function.  It is the 'remaining balance on loan' which might be useful as
        it's own function, but is easily calculated with the 'fv' function.
        """
        return npf.fv(rate, (per - 1), pmt, pv, when)

    @staticmethod
    def ppmt(rate, per, nper, pv, fv=0, when='end'):
        """
        Compute the payment against loan principal.

        Parameters
        ----------
        rate : array_like
            Rate of interest (per period)
        per : array_like, int
            Amount paid against the loan changes.  The `per` is the period of
            interest.
        nper : array_like
            Number of compounding periods
        pv : array_like
            Present value
        fv : array_like, optional
            Future value
        when : {{'begin', 1}, {'end', 0}}, {string, int}
            When payments are due ('begin' (1) or 'end' (0))

        See Also
        --------
        pmt, pv, ipmt

        """
        total = npf.pmt(rate, nper, pv, fv, when)
        return total - npf.ipmt(rate, per, nper, pv, fv, when)

    @staticmethod
    def pv(rate, nper, pmt, fv=0, when='end'):
        """
        Compute the present value.

        Given:
         * a future value, `fv`
         * an interest `rate` compounded once per period, of which
           there are
         * `nper` total
         * a (fixed) payment, `pmt`, paid either
         * at the beginning (`when` = {'begin', 1}) or the end
           (`when` = {'end', 0}) of each period

        Return:
           the value now

        Parameters
        ----------
        rate : array_like
            Rate of interest (per period)
        nper : array_like
            Number of compounding periods
        pmt : array_like
            Payment
        fv : array_like, optional
            Future value
        when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
            When payments are due ('begin' (1) or 'end' (0))

        Returns
        -------
        out : ndarray, float
            Present value of a series of payments or investments.

        Notes
        -----
        The present value is computed by solving the equation::

         fv +
         pv*(1 + rate)**nper +
         pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) = 0

        or, when ``rate = 0``::

         fv + pv + pmt * nper = 0

        for `pv`, which is then returned.

        References
        ----------
        .. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
           Open Document Format for Office Applications (OpenDocument)v1.2,
           Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
           Pre-Draft 12. Organization for the Advancement of Structured Information
           Standards (OASIS). Billerica, MA, USA. [ODT Document].
           Available:
           http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
           OpenDocument-formula-20090508.odt

        Examples
        --------
        >>> import numpy as np
        >>> import numpy_financial as npf

        What is the present value (e.g., the initial investment)
        of an investment that needs to total $15692.93
        after 10 years of saving $100 every month?  Assume the
        interest rate is 5% (annually) compounded monthly.

        >>> npf.pv(0.05/12, 10*12, -100, 15692.93)
        -100.00067131625819

        By convention, the negative sign represents cash flow out
        (i.e., money not available today).  Thus, to end up with
        $15,692.93 in 10 years saving $100 a month at 5% annual
        interest, one's initial deposit should also be $100.

        If any input is array_like, ``pv`` returns an array of equal shape.
        Let's compare different interest rates in the example above:

        >>> a = np.array((0.05, 0.04, 0.03))/12
        >>> npf.pv(a, 10*12, -100, 15692.93)
        array([ -100.00067132,  -649.26771385, -1273.78633713]) # may vary

        So, to end up with the same $15692.93 under the same $100 per month
        "savings plan," for annual interest rates of 4% and 3%, one would
        need initial investments of $649.27 and $1273.79, respectively.

        """
        when = npf._convert_when(when)
        (rate, nper, pmt, fv, when) = map(np.asarray, [rate, nper, pmt, fv, when])
        temp = (1+rate)**nper
        fact = np.where(rate == 0, nper, (1+rate*when)*(temp-1)/rate)
        return -(fv + pmt*fact)/temp

    # Computed with Sage
    #  (y + (r + 1)^n*x + p*((r + 1)^n - 1)*(r*w + 1)/r)/(n*(r + 1)^(n - 1)*x -
    #  p*((r + 1)^n - 1)*(r*w + 1)/r^2 + n*p*(r + 1)^(n - 1)*(r*w + 1)/r +
    #  p*((r + 1)^n - 1)*w/r)

    @staticmethod
    def _g_div_gp(r, n, p, x, y, w):
        # Evaluate g(r_n)/g'(r_n), where g =
        # fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1)
        t1 = (r+1)**n
        t2 = (r+1)**(n-1)
        g = y + t1*x + p*(t1 - 1) * (r*w + 1) / r
        gp = (n*t2*x
              - p*(t1 - 1) * (r*w + 1) / (r**2)
              + n*p*t2 * (r*w + 1) / r
              + p*(t1 - 1) * w/r)
        return g / gp


    # Use Newton's iteration until the change is less than 1e-6
    #  for all values or a maximum of 100 iterations is reached.
    #  Newton's rule is
    #  r_{n+1} = r_{n} - g(r_n)/g'(r_n)
    #     where
    #  g(r) is the formula
    #  g'(r) is the derivative with respect to r.

    @staticmethod
    def rate(nper, pmt, pv, fv, when='end', guess=None, tol=None, maxiter=100):
        """
        Compute the rate of interest per period.

        Parameters
        ----------
        nper : array_like
            Number of compounding periods
        pmt : array_like
            Payment
        pv : array_like
            Present value
        fv : array_like
            Future value
        when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
            When payments are due ('begin' (1) or 'end' (0))
        guess : Number, optional
            Starting guess for solving the rate of interest, default 0.1
        tol : Number, optional
            Required tolerance for the solution, default 1e-6
        maxiter : int, optional
            Maximum iterations in finding the solution

        Notes
        -----
        The rate of interest is computed by iteratively solving the
        (non-linear) equation::

         fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1) = 0

        for ``rate``.

        References
        ----------
        Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May). Open Document
        Format for Office Applications (OpenDocument)v1.2, Part 2: Recalculated
        Formula (OpenFormula) Format - Annotated Version, Pre-Draft 12.
        Organization for the Advancement of Structured Information Standards
        (OASIS). Billerica, MA, USA. [ODT Document]. Available:
        http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
        OpenDocument-formula-20090508.odt

        """
        when = npf._convert_when(when)
        default_type = Decimal if isinstance(pmt, Decimal) else float

        # Handle casting defaults to Decimal if/when pmt is a Decimal and
        # guess and/or tol are not given default values
        if guess is None:
            guess = default_type('0.1')

        if tol is None:
            tol = default_type('1e-6')

        (nper, pmt, pv, fv, when) = map(np.asarray, [nper, pmt, pv, fv, when])

        rn = guess
        iterator = 0
        close = False
        while (iterator < maxiter) and not close:
            rnp1 = rn - npf._g_div_gp(rn, nper, pmt, pv, fv, when)
            diff = abs(rnp1-rn)
            close = np.all(diff < tol)
            iterator += 1
            rn = rnp1
        if not close:
            # Return nan's in array of the same shape as rn
            return default_type(np.nan) + rn
        else:
            return rn

    @staticmethod
    def _roots(p):
        """Modified version of NumPy's roots function.

        NumPy's roots uses the companion matrix method, which divides by
        p[0]. This can causes overflows/underflows. Instead form a
        modified companion matrix that is scaled by 2^c * p[0], where the
        exponent c is chosen to balance the magnitudes of the
        coefficients. Since scaling the matrix just scales the
        eigenvalues, we can remove the scaling at the end.

        Scaling by a power of 2 is chosen to avoid rounding errors.

        """
        _, e = np.frexp(p)
        # Balance the most extreme exponents e_max and e_min by solving
        # the equation
        #
        # |c + e_max| = |c + e_min|.
        #
        # Round the exponent to an integer to avoid rounding errors.
        c = int(-0.5 * (np.max(e) + np.min(e)))
        p = np.ldexp(p, c)

        A = np.diag(np.full(p.size - 2, p[0]), k=-1)
        A[0,:] = -p[1:]
        eigenvalues = np.linalg.eigvals(A)
        return eigenvalues / p[0]

    @staticmethod
    def irr(values):
        """
        Return the Internal Rate of Return (IRR).

        This is the "average" periodically compounded rate of return
        that gives a net present value of 0.0; for a more complete explanation,
        see Notes below.

        :class:`decimal.Decimal` type is not supported.

        Parameters
        ----------
        values : array_like, shape(N,)
            Input cash flows per time period.  By convention, net "deposits"
            are negative and net "withdrawals" are positive.  Thus, for
            example, at least the first element of `values`, which represents
            the initial investment, will typically be negative.

        Returns
        -------
        out : float
            Internal Rate of Return for periodic input values.

        Notes
        -----
        The IRR is perhaps best understood through an example (illustrated
        using np.irr in the Examples section below).  Suppose one invests 100
        units and then makes the following withdrawals at regular (fixed)
        intervals: 39, 59, 55, 20.  Assuming the ending value is 0, one's 100
        unit investment yields 173 units; however, due to the combination of
        compounding and the periodic withdrawals, the "average" rate of return
        is neither simply 0.73/4 nor (1.73)^0.25-1.  Rather, it is the solution
        (for :math:`r`) of the equation:

        .. math:: -100 + \\frac{39}{1+r} + \\frac{59}{(1+r)^2}
         + \\frac{55}{(1+r)^3} + \\frac{20}{(1+r)^4} = 0

        In general, for `values` :math:`= [v_0, v_1, ... v_M]`,
        irr is the solution of the equation: [G]_

        .. math:: \\sum_{t=0}^M{\\frac{v_t}{(1+irr)^{t}}} = 0

        References
        ----------
        .. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
           Addison-Wesley, 2003, pg. 348.

        Examples
        --------
        >>> import numpy_financial as npf

        >>> round(npf.irr([-100, 39, 59, 55, 20]), 5)
        0.28095
        >>> round(npf.irr([-100, 0, 0, 74]), 5)
        -0.0955
        >>> round(npf.irr([-100, 100, 0, -7]), 5)
        -0.0833
        >>> round(npf.irr([-100, 100, 0, 7]), 5)
        0.06206
        >>> round(npf.irr([-5, 10.5, 1, -8, 1]), 5)
        0.0886

        """
        values = np.atleast_1d(values)
        if values.ndim != 1:
            raise ValueError("Cashflows must be a rank-1 array")

        # Strip leading and trailing zeros. Since we only care about
        # positive roots we can neglect roots at zero.
        non_zero = np.nonzero(np.ravel(values))[0]
        values = values[int(non_zero[0]):int(non_zero[-1])+1]

        res = npf._roots(values[::-1])

        mask = (res.imag == 0) & (res.real > 0)
        if not mask.any():
            return np.nan
        res = res[mask].real
        # NPV(rate) = 0 can have more than one solution so we return
        # only the solution closest to zero.
        rate = 1/res - 1
        rate = rate.item(np.argmin(np.abs(rate)))
        return rate

    @staticmethod
    def npv(rate, values):
        """
        Returns the NPV (Net Present Value) of a cash flow series.

        Parameters
        ----------
        rate : scalar
            The discount rate.
        values : array_like, shape(M, )
            The values of the time series of cash flows.  The (fixed) time
            interval between cash flow "events" must be the same as that for
            which `rate` is given (i.e., if `rate` is per year, then precisely
            a year is understood to elapse between each cash flow event).  By
            convention, investments or "deposits" are negative, income or
            "withdrawals" are positive; `values` must begin with the initial
            investment, thus `values[0]` will typically be negative.

        Returns
        -------
        out : float
            The NPV of the input cash flow series `values` at the discount
            `rate`.

        Warnings
        --------
        ``npv`` considers a series of cashflows starting in the present (t = 0).
        NPV can also be defined with a series of future cashflows, paid at the
        end, rather than the start, of each period. If future cashflows are used,
        the first cashflow `values[0]` must be zeroed and added to the net
        present value of the future cashflows. This is demonstrated in the
        examples.

        Notes
        -----
        Returns the result of: [G]_

        .. math :: \\sum_{t=0}^{M-1}{\\frac{values_t}{(1+rate)^{t}}}

        References
        ----------
        .. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
           Addison-Wesley, 2003, pg. 346.

        Examples
        --------
        >>> import numpy as np
        >>> import numpy_financial as npf

        Consider a potential project with an initial investment of $40 000 and
        projected cashflows of $5 000, $8 000, $12 000 and $30 000 at the end of
        each period discounted at a rate of 8% per period. To find the project's
        net present value:

        >>> rate, cashflows = 0.08, [-40_000, 5_000, 8_000, 12_000, 30_000]
        >>> npf.npv(rate, cashflows).round(5)
        3065.22267

        It may be preferable to split the projected cashflow into an initial
        investment and expected future cashflows. In this case, the value of
        the initial cashflow is zero and the initial investment is later added
        to the future cashflows net present value:

        >>> initial_cashflow = cashflows[0]
        >>> cashflows[0] = 0
        >>> np.round(npf.npv(rate, cashflows) + initial_cashflow, 5)
        3065.22267

        """
        values = np.asarray(values)
        return (values / (1+rate)**np.arange(0, len(values))).sum(axis=0)

    @staticmethod
    def mirr(values, finance_rate, reinvest_rate):
        """
        Modified internal rate of return.

        Parameters
        ----------
        values : array_like
            Cash flows (must contain at least one positive and one negative
            value) or nan is returned.  The first value is considered a sunk
            cost at time zero.
        finance_rate : scalar
            Interest rate paid on the cash flows
        reinvest_rate : scalar
            Interest rate received on the cash flows upon reinvestment

        Returns
        -------
        out : float
            Modified internal rate of return

        """
        values = np.asarray(values)
        n = values.size

        # Without this explicit cast the 1/(n - 1) computation below
        # becomes a float, which causes TypeError when using Decimal
        # values.
        if isinstance(finance_rate, Decimal):
            n = Decimal(n)

        pos = values > 0
        neg = values < 0
        if not (pos.any() and neg.any()):
            return np.nan
        numer = np.abs(npf.npv(reinvest_rate, values*pos))
        denom = np.abs(npf.npv(finance_rate, values*neg))
        return (numer/denom)**(1/(n - 1))*(1 + reinvest_rate) - 1
